Here, all spaces are Banach spaces.
Definition: A map $S:X \to X$ is compact if for every bounded sequence $\{u_n\}$, there exists a subsequence $\{u_{n_k}\}$ such that $\{S(u_{n_k})\}$ converges in $X$.
Question: suppose $A$ is compactly embedded in $B$. Suppose a map $T:A \to A$ is continuous. Is there any chance that $T:A \to A$ is compact? ($T:B \to A$ is not definable or is ill-defined). If context is important: take $A=C^{2, \alpha}$ and $B=C^{0, \alpha}$, Hölder continuous functions. It is true that $A$ is compactly embedded in $B$ (the norms are different on $A$ and $B$ — they are the standard norms on Wikipedia).
Thoughts: I don't think so in general. I can't see any way, unless there's some cool theorem I'm not aware of (and I'm not aware of a lot of things so maybe this is possible).
Motivation: want to show existence to a PDE problem.
Best Answer
If the embedding $e: A \to B$ is compact, and there is a continuous linear map $S:B \to A$ such that $T$ equals the composite $S \circ e$, then $T$ will be compact (since the composite of a compact operator and any other continuous linear map is always compact).
[This idea is used in elliptic PDE theory, to deduce compactness of the inverse of elliptic differential operators. The point (stated very roughly) is that applying the inverse of an elliptic operator should improve the differentiability class of a function.]
Without such a factorization, it's not clear how you would ever hope to link the behaviour of $T$ and the compactness of the embedding $e$.